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301. Which of the following elements most commonly exhibits variable valency?
ⓐ. Sodium
ⓑ. Magnesium
ⓒ. Iron
ⓓ. Neon
Correct Answer: Iron
Explanation: Iron (Fe) is a transition metal that commonly shows +2 and +3 oxidation states. This is due to the close energy of its 3d and 4s orbitals, allowing variable numbers of electrons to participate in bonding.
302. What are the two common oxidation states of copper?
ⓐ. +1 and +2
ⓑ. +2 and +3
ⓒ. +3 and +4
ⓓ. +1 and +3
Correct Answer: +1 and +2
Explanation: Copper (Cu) belongs to Group 11. It commonly shows +1 (Cu₂O) and +2 (CuO, CuSO₄) oxidation states. The stability of these states depends on lattice energy and ionization enthalpies.
303. Which element shows variable valencies of +2 and +4?
ⓐ. Carbon
ⓑ. Lead
ⓒ. Sodium
ⓓ. Fluorine
Correct Answer: Lead
Explanation: Lead (Pb, Group 14) exhibits +2 and +4 oxidation states. +2 is more stable in heavier elements due to the inert pair effect (reluctance of ns² electrons to participate in bonding).
304. Which oxidation states are commonly shown by manganese?
ⓐ. +2, +3, +4, +6, +7
ⓑ. +1 and +2
ⓒ. +2 and +3 only
ⓓ. +3 and +5 only
Correct Answer: +2, +3, +4, +6, +7
Explanation: Manganese (Mn, Z = 25) has a wide range of oxidation states due to availability of 3d and 4s electrons. Examples: MnCl₂ (+2), Mn₂O₃ (+3), MnO₂ (+4), KMnO₄ (+7).
305. Which of the following is NOT an example of variable valency?
ⓐ. Fe → +2, +3
ⓑ. Cu → +1, +2
ⓒ. Na → +1
ⓓ. Sn → +2, +4
Correct Answer: Na → +1
Explanation: Sodium (Group 1) has a fixed valency of +1, as it loses only one ns¹ electron. Variable valency is typical of transition metals and heavier p-block elements like Sn, Pb.
306. Which oxidation states are commonly shown by tin (Sn)?
ⓐ. +1 and +3
ⓑ. +2 and +4
ⓒ. +2 and +3
ⓓ. +3 and +5
Correct Answer: +2 and +4
Explanation: Tin (Sn, Group 14) exhibits +2 (SnCl₂) and +4 (SnCl₄). Like lead, the +2 state is stabilized by the inert pair effect, but +4 is also stable in lighter compounds.
307. Why do transition elements often exhibit variable valency?
ⓐ. Because their s and d orbitals have similar energies
ⓑ. Because they have unstable nuclei
ⓒ. Because they always form covalent bonds
ⓓ. Because they have very large atomic radii
Correct Answer: Because their s and d orbitals have similar energies
Explanation: In transition metals, ns and (n–1)d orbitals are close in energy, so electrons from both shells can take part in bonding. This leads to multiple stable oxidation states, unlike s-block elements.
308. Which element shows +2 and +3 oxidation states, with +3 being more stable?
ⓐ. Iron
ⓑ. Calcium
ⓒ. Sodium
ⓓ. Potassium
Correct Answer: Iron
Explanation: Iron (Fe) shows +2 (FeO) and +3 (Fe₂O₃). The +3 state is generally more stable due to higher lattice energy and hydration enthalpy of Fe³⁺.
309. What are the common oxidation states of chromium?
ⓐ. +2, +3, +6
ⓑ. +1, +2
ⓒ. +2 only
ⓓ. +4 and +5 only
Correct Answer: +2, +3, +6
Explanation: Chromium (Cr, Z = 24) exhibits +2 (CrCl₂), +3 (Cr₂O₃), and +6 (K₂Cr₂O₇). The +6 state is strongly oxidizing, while +3 is the most stable oxidation state in aqueous chemistry.
310. Which heavy p-block element exhibits variable valencies of +3 and +5?
ⓐ. Phosphorus
ⓑ. Chlorine
ⓒ. Bismuth
ⓓ. Sulfur
Correct Answer: Bismuth
Explanation: Bismuth (Bi, Group 15) shows +3 and +5 states, but the +3 state is more stable due to the inert pair effect. This is typical of heavier Group 15 elements where ns² electrons resist bonding participation.
311. Why do second-period elements often show anomalous properties compared to heavier congeners?
ⓐ. Because they have unstable nuclei
ⓑ. Because they have small size, high electronegativity, and absence of d-orbitals
ⓒ. Because they belong to transition metals
ⓓ. Because they always form cations
Correct Answer: Because they have small size, high electronegativity, and absence of d-orbitals
Explanation: Elements like Li, Be, B, C, N, O, and F (second period) deviate in properties from their group members due to their small atomic radius, high electronegativity, and lack of vacant d-orbitals for bonding, which restricts their coordination numbers and oxidation states.
312. Why can’t second-period elements expand their octet?
ⓐ. Because they are highly reactive
ⓑ. Because they have no vacant d-orbitals in their valence shell
ⓒ. Because they have lower atomic mass
ⓓ. Because they are non-metals
Correct Answer: Because they have no vacant d-orbitals in their valence shell
Explanation: The maximum valence shell for second-period elements is n=2, which includes only s and p orbitals. Without d-orbitals, they cannot expand their octet beyond 8 electrons. For example, carbon never exceeds valency 4.
313. Why is fluorine the strongest oxidizing agent among the halogens?
ⓐ. It has low electronegativity
ⓑ. It has high electronegativity and small size
ⓒ. It has vacant d-orbitals
ⓓ. It has very high atomic radius
Correct Answer: It has high electronegativity and small size
Explanation: Fluorine (χ = 3.98) has the highest electronegativity and smallest size among halogens. These properties allow it to attract electrons most effectively, making it the strongest oxidizing agent.
314. Why is lithium different from other alkali metals?
ⓐ. Because of its larger radius and higher mass
ⓑ. Because of its small size, high polarizing power, and strong covalent bond formation
ⓒ. Because it is heavier than sodium
ⓓ. Because it has more d-orbitals
Correct Answer: Because of its small size, high polarizing power, and strong covalent bond formation
Explanation: Lithium forms covalent compounds (LiCl, Li₂O) unlike other alkali metals which form ionic compounds. Its small radius and high charge density give it strong polarizing ability, leading to covalency.
315. Why does beryllium form covalent compounds instead of ionic ones like other Group 2 metals?
ⓐ. Because it has very low ionization enthalpy
ⓑ. Because of its small size and high ionization enthalpy
ⓒ. Because it has vacant d-orbitals
ⓓ. Because it is radioactive
Correct Answer: Because of its small size and high ionization enthalpy
Explanation: Beryllium has a small radius and high ionization enthalpy, so it cannot easily lose 2 electrons to form Be²⁺. Instead, it forms covalent compounds (BeCl₂, BeH₂) by sharing electrons.
316. Why do second-period elements form strong π-bonds more easily than heavier elements?
ⓐ. Because of their large atomic size
ⓑ. Because of effective side-by-side overlap of small p-orbitals
ⓒ. Because of the presence of d-orbitals
ⓓ. Because they are metallic in nature
Correct Answer: Because of effective side-by-side overlap of small p-orbitals
Explanation: Elements like C, N, and O in the second period have small p-orbitals that overlap effectively to form strong π-bonds (C=C, N≡N, C=O). Heavier congeners with larger orbitals show weaker π-bonding.
317. Why does nitrogen not form pentahalides like phosphorus?
ⓐ. Because nitrogen is less reactive
ⓑ. Because nitrogen lacks d-orbitals to expand its octet
ⓒ. Because nitrogen has low electronegativity
ⓓ. Because nitrogen is metallic in nature
Correct Answer: Because nitrogen lacks d-orbitals to expand its octet
Explanation: Nitrogen can form only up to four bonds (NH₄⁺) due to the absence of 2d orbitals. Phosphorus, with 3d orbitals available, can expand its octet to form PF₅.
318. Why is oxygen able to form multiple bonds while sulfur prefers single bonds?
ⓐ. Because oxygen has low electronegativity
ⓑ. Because oxygen has small size and effective p-orbital overlap
ⓒ. Because sulfur has no d-orbitals
ⓓ. Because sulfur is a metal
Correct Answer: Because oxygen has small size and effective p-orbital overlap
Explanation: Oxygen (2p orbitals) forms strong π-bonds (O=O, C=O) due to efficient overlap. Sulfur, being larger, cannot overlap effectively, so it prefers single bonds or hypervalent compounds (SF₆).
319. Why do second-period elements form stronger hydrogen bonds than heavier elements?
ⓐ. Because they are less electronegative
ⓑ. Because of their small size and high electronegativity
ⓒ. Because they have vacant d-orbitals
ⓓ. Because they are metals
Correct Answer: Because of their small size and high electronegativity
Explanation: Elements like O, N, and F form strong hydrogen bonds due to their compact size and strong attraction for electrons. Larger congeners (S, P, Cl) form much weaker or negligible H-bonds.
320. Why does carbon form stable multiple bonds, but silicon prefers single bonds?
ⓐ. Because silicon is more electronegative
ⓑ. Because carbon is smaller, allowing effective p–p overlap
ⓒ. Because silicon lacks p-orbitals
ⓓ. Because carbon has d-orbitals
Correct Answer: Because carbon is smaller, allowing effective p–p overlap
Explanation: Carbon’s small 2p orbitals overlap effectively, forming stable C=C and C≡C bonds. Silicon’s 3p orbitals are more diffuse, giving poor overlap, so it prefers single bonds like Si–Si or Si–O.
321. What is meant by the diagonal relationship in the periodic table?
ⓐ. Similarities between elements in the same group
ⓑ. Similarities between elements in the same period
ⓒ. Similarities between diagonally adjacent elements of Period 2 and Period 3
ⓓ. Differences between elements of the same block
Correct Answer: Similarities between diagonally adjacent elements of Period 2 and Period 3
Explanation: Diagonal relationship occurs because elements of Period 2 (Li, Be, B) show properties similar to Period 3 elements placed diagonally below them (Mg, Al, Si). This arises due to comparable atomic size, charge/radius ratio, and electronegativity.
322. Which pair of elements shows diagonal relationship in the periodic table?
ⓐ. Li and Na
ⓑ. Be and Mg
ⓒ. Li and Mg
ⓓ. B and Al
Correct Answer: Li and Mg
Explanation: Lithium (Period 2, Group 1) and Magnesium (Period 3, Group 2) show diagonal relationship. Both form nitrides (Li₃N, Mg₃N₂), carbonates that decompose on heating, and soluble halides, unlike other alkali metals and alkaline earth metals.
323. Which of the following is a similarity between Li and Mg?
ⓐ. Both form stable bicarbonates
ⓑ. Both form nitrides directly with nitrogen
ⓒ. Both form soluble carbonates like Na₂CO₃
ⓓ. Both are highly reactive like Cs
Correct Answer: Both form nitrides directly with nitrogen
Explanation: Lithium reacts with nitrogen to form Li₃N, and magnesium reacts with nitrogen to form Mg₃N₂. Other alkali metals do not form stable nitrides, showing the diagonal similarity between Li and Mg.
324. Which property is common to both Be and Al due to diagonal relationship?
ⓐ. Formation of strong basic hydroxides
ⓑ. Formation of amphoteric oxides and hydroxides
ⓒ. Formation of stable +1 oxidation state
ⓓ. High solubility of carbonates
Correct Answer: Formation of amphoteric oxides and hydroxides
Explanation: BeO and Al₂O₃ are amphoteric, reacting with both acids and bases. This similarity is explained by their comparable charge density (high polarizing power), which is typical in diagonal relationship.
325. Which pair shows diagonal similarity in forming covalent halides?
ⓐ. Na and K
ⓑ. Li and Mg
ⓒ. Be and Mg
ⓓ. Ca and Sr
Correct Answer: Li and Mg
Explanation: Both lithium and magnesium form covalent halides (LiCl, MgCl₂) that are soluble in organic solvents. Other alkali metals and alkaline earth metals usually form ionic halides.
326. Which property is shared by boron and silicon due to diagonal relationship?
ⓐ. Both form basic oxides
ⓑ. Both form volatile hydrides like BH₃ and SiH₄
ⓒ. Both form strong ionic halides
ⓓ. Both show metallic character
Correct Answer: Both form volatile hydrides like BH₃ and SiH₄
Explanation: Boron forms covalent hydrides (boranes, e.g., BH₃), while silicon forms hydrides (silanes, e.g., SiH₄). These are covalent, volatile compounds, unlike ionic hydrides formed by many other group members.
327. Which is a similarity between Be and Al compounds?
ⓐ. Both form highly soluble carbonates
ⓑ. Both form covalent chlorides that hydrolyze in water
ⓒ. Both form simple ionic oxides
ⓓ. Both form halides that are inert
Correct Answer: Both form covalent chlorides that hydrolyze in water
Explanation: BeCl₂ and AlCl₃ are covalent and easily hydrolyze in water, forming hydroxides and releasing HCl. This is unlike MgCl₂, which is ionic and stable in water.
328. Which property shows similarity between B and Si?
ⓐ. Both form oxides that are acidic in nature
ⓑ. Both form strongly basic oxides
ⓒ. Both form stable +1 oxidation states
ⓓ. Both are good conductors of electricity
Correct Answer: Both form oxides that are acidic in nature
Explanation: Boron forms B₂O₃ and silicon forms SiO₂. Both oxides dissolve in alkalis to form borates and silicates, showing acidic nature. This diagonal similarity is not observed with Al or C oxides.
329. Why does the diagonal relationship exist in the periodic table?
ⓐ. Because atomic masses are the same diagonally
ⓑ. Because of balancing effects of atomic size and electronegativity
ⓒ. Because diagonal elements belong to the same block
ⓓ. Because they have identical valence configurations
Correct Answer: Because of balancing effects of atomic size and electronegativity
Explanation: Down a group, atomic size increases and electronegativity decreases. Across a period, atomic size decreases and electronegativity increases. Along a diagonal, these effects cancel out, giving elements similar charge density and chemical behavior.
330. Which of the following pairs shows amphoteric oxide formation due to diagonal relationship?
ⓐ. Li and Na
ⓑ. Be and Al
ⓒ. B and C
ⓓ. Mg and Ca
Correct Answer: Be and Al
Explanation: BeO and Al₂O₃ are amphoteric, reacting with both acids and bases. This is not true for Na₂O or CaO (which are basic). Thus, Be and Al show diagonal similarity in amphoteric behavior.
331. Why does lithium show different chemistry compared to other alkali metals?
ⓐ. Because lithium has the largest atomic radius
ⓑ. Because lithium has very low ionization enthalpy
ⓒ. Because lithium has small size, high polarizing power, and forms covalent compounds
ⓓ. Because lithium has vacant d-orbitals
Correct Answer: Because lithium has small size, high polarizing power, and forms covalent compounds
Explanation: Lithium differs from Na, K, Rb, Cs because of its small ionic radius and high charge density. Li⁺ strongly polarizes anions, forming covalent compounds (LiCl soluble in organic solvents). Other alkali metals mainly form ionic compounds.
332. Why does beryllium behave differently from other alkaline earth metals?
ⓐ. Because Be is the heaviest element in Group 2
ⓑ. Because Be²⁺ is large and ionic
ⓒ. Because Be²⁺ is very small and highly polarizing, leading to covalent bond formation
ⓓ. Because Be has extra f-orbitals
Correct Answer: Because Be²⁺ is very small and highly polarizing, leading to covalent bond formation
Explanation: Beryllium forms covalent compounds (BeCl₂, BeH₂) unlike other Group 2 metals. Its oxides/hydroxides (BeO, Be(OH)₂) are amphoteric, while others are basic. This arises from its small size and high ionization enthalpy.
333. Why does boron show chemistry very different from aluminum?
ⓐ. Because boron is metallic, aluminum is non-metallic
ⓑ. Because boron has very high ionization enthalpy, small size, and no d-orbitals
ⓒ. Because boron easily expands its octet
ⓓ. Because aluminum has fewer valence electrons
Correct Answer: Because boron has very high ionization enthalpy, small size, and no d-orbitals
Explanation: Boron is strictly non-metallic, forming only covalent compounds (e.g., BF₃). It does not form simple ionic salts like Al³⁺. Aluminum, being larger and having vacant d-orbitals, shows more metallic and amphoteric behavior.
334. Which of the following is a unique property of lithium compared to other alkali metals?
ⓐ. Lithium forms a stable nitride (Li₃N) directly with nitrogen
ⓑ. Lithium has the lowest melting point among alkali metals
ⓒ. Lithium forms only ionic halides
ⓓ. Lithium is highly reactive with noble gases
Correct Answer: Lithium forms a stable nitride (Li₃N) directly with nitrogen
Explanation: Lithium is the only alkali metal that reacts directly with nitrogen to form Li₃N. Other alkali metals do not form stable nitrides under normal conditions.
335. Which property distinguishes beryllium hydroxide from other Group 2 hydroxides?
ⓐ. It is strongly basic
ⓑ. It is strongly acidic
ⓒ. It is amphoteric
ⓓ. It is insoluble in water
Correct Answer: It is amphoteric
Explanation: Be(OH)₂ reacts with both acids and bases, unlike Mg(OH)₂, Ca(OH)₂, which are basic. This amphoteric behavior is due to Be’s small size and high polarizing ability.
336. Why is boron oxide (B₂O₃) acidic while aluminum oxide (Al₂O₃) is amphoteric?
ⓐ. Because boron has lower electronegativity
ⓑ. Because boron cannot expand its octet, while aluminum can use d-orbitals
ⓒ. Because aluminum has a smaller atomic size
ⓓ. Because boron is metallic
Correct Answer: Because boron cannot expand its octet, while aluminum can use d-orbitals
Explanation: B₂O₃ dissolves in alkalis to form borates, showing acidic nature. Al₂O₃ reacts with both acids and bases, showing amphoteric nature due to availability of d-orbitals and larger size.
337. Why does lithium carbonate decompose easily compared to sodium carbonate?
ⓐ. Because lithium carbonate is more basic
ⓑ. Because Li⁺ ion has high polarizing power, destabilizing CO₃²⁻
ⓒ. Because sodium has higher ionization enthalpy
ⓓ. Because Li is heavier than Na
Correct Answer: Because Li⁺ ion has high polarizing power, destabilizing CO₃²⁻
Explanation: Li⁺ is very small and highly polarizing, distorting the carbonate ion and making Li₂CO₃ unstable to heat. It decomposes into Li₂O and CO₂. Na₂CO₃ is stable to heat.
338. Why is BeCl₂ covalent while MgCl₂ is ionic?
ⓐ. Because Be has a larger radius
ⓑ. Because Be²⁺ has high polarizing power and forms covalent bonds
ⓒ. Because Mg²⁺ has lower charge density
ⓓ. Because Be has a lower ionization enthalpy
Correct Answer: Because Be²⁺ has high polarizing power and forms covalent bonds
Explanation: According to Fajan’s rule, Be²⁺ (small, highly charged) polarizes Cl⁻ strongly, leading to covalent bonding. Mg²⁺ is larger and less polarizing, so MgCl₂ is ionic.
339. Why does boron not form simple ionic compounds like BCl₃⁻?
ⓐ. Because boron is metallic
ⓑ. Because boron has low electronegativity
ⓒ. Because boron is small, has high ionization enthalpy, and forms covalent bonds instead
ⓓ. Because boron has vacant d-orbitals
Correct Answer: Because boron is small, has high ionization enthalpy, and forms covalent bonds instead
Explanation: Boron cannot lose 3 electrons easily due to high ionization enthalpy. Instead, it shares electrons to form covalent compounds like BF₃ and BCl₃.
340. Why does lithium resemble magnesium more than sodium?
ⓐ. Because lithium is larger in size
ⓑ. Because both lithium and magnesium form covalent compounds and carbonates that decompose on heating
ⓒ. Because both are transition elements
ⓓ. Because sodium belongs to a different period
Correct Answer: Because both lithium and magnesium form covalent compounds and carbonates that decompose on heating
Explanation: Lithium and magnesium (diagonal relationship) form soluble halides, unstable carbonates, and nitrides. Sodium differs by forming stable ionic compounds. This makes Li closer to Mg in chemical behavior.
341. Who is credited with introducing the concept of “atomic number” as the basis for classification?
ⓐ. Dmitri Mendeleev
ⓑ. Henry Moseley
ⓒ. J.J. Thomson
ⓓ. John Dalton
Correct Answer: Henry Moseley
Explanation: In 1913, Moseley showed that atomic number (Z), not atomic mass, is the fundamental property for classification. His X-ray studies confirmed that the periodic properties are periodic functions of atomic number, leading to the modern periodic law.
342. Which property is not periodic in nature?
ⓐ. Atomic radius
ⓑ. Ionization enthalpy
ⓒ. Electronegativity
ⓓ. Neutron number
Correct Answer: Neutron number
Explanation: Atomic radius, ionization enthalpy, and electronegativity vary periodically with atomic number. Neutron number, however, does not show a periodic pattern, since isotopes of the same element have different neutrons.
343. Which element is placed in Group 1 but also shows resemblance with Group 17 elements?
ⓐ. Helium
ⓑ. Hydrogen
ⓒ. Lithium
ⓓ. Sodium
Correct Answer: Hydrogen
Explanation: Hydrogen has ns¹ configuration like alkali metals (Group 1), but it also forms diatomic molecules (H₂) and gains one electron like halogens (Group 17). Its dual nature makes its position unique.
344. Which statement about d-block elements is correct?
ⓐ. They all have fully filled d-subshells in the ground state
ⓑ. They exhibit variable oxidation states
ⓒ. They never form colored compounds
ⓓ. They all behave like alkali metals
Correct Answer: They exhibit variable oxidation states
Explanation: d-block elements (transition metals) have ns and (n–1)d orbitals close in energy, allowing different numbers of electrons to participate in bonding. This leads to variable oxidation states (e.g., Fe²⁺, Fe³⁺).
345. What is the periodic trend of metallic character across a period?
ⓐ. Increases from left to right
ⓑ. Decreases from left to right
ⓒ. Remains constant
ⓓ. Fluctuates irregularly
Correct Answer: Decreases from left to right
Explanation: Metallic character depends on ease of losing electrons. Across a period, ionization enthalpy increases due to smaller size and higher nuclear charge, so metallic character decreases.
346. Which of the following is true for noble gases?
ⓐ. They have the smallest atomic radii in their periods (except He with van der Waals radius)
ⓑ. They have the largest atomic radii in their periods
ⓒ. They have the most negative electron gain enthalpies
ⓓ. They are highly electronegative
Correct Answer: They have the smallest atomic radii in their periods (except He with van der Waals radius)
Explanation: Noble gases are inert and do not form covalent radii, but their van der Waals radii are large. However, when compared based on effective nuclear attraction, they are considered among the smallest in covalent/atomic radius terms in their periods.
347. Which of the following pairs is an isoelectronic series?
ⓐ. O²⁻, F⁻, Ne, Na⁺
ⓑ. Li⁺, Be²⁺, B³⁺, C⁴⁺
ⓒ. Cl⁻, Ar, K⁺, Ca²⁺
ⓓ. All of the above
Correct Answer: All of the above
Explanation: Isoelectronic species have equal numbers of electrons. Each option contains sets of ions/atoms with the same number of electrons but different nuclear charges, showing systematic size trends.
348. Which property best explains the gradual change from metallic to non-metallic behavior across a period?
ⓐ. Increase in ionization enthalpy and electronegativity
ⓑ. Increase in atomic radius
ⓒ. Decrease in nuclear charge
ⓓ. Increase in number of isotopes
Correct Answer: Increase in ionization enthalpy and electronegativity
Explanation: Metals lose electrons, while non-metals gain electrons. Across a period, ionization enthalpy rises and electronegativity increases, shifting character from metallic (left) to non-metallic (right).
349. What is the position of lanthanides and actinides in the long-form periodic table?
ⓐ. In Groups 3 and 4
ⓑ. In Periods 6 and 7, placed separately below the main table
ⓒ. Between Group 1 and Group 2
ⓓ. In Group 18
Correct Answer: In Periods 6 and 7, placed separately below the main table
Explanation: Lanthanides (Z=57–71) and actinides (Z=89–103) belong to the f-block. To keep the periodic table compact, they are shown as two separate rows below the main body, though they actually fit into Periods 6 and 7 under Group 3.
350. Which is a unique property of second-period elements not observed in heavier congeners?
ⓐ. Formation of multiple bonds (C=C, N≡N, C=O) due to small size and strong p–p overlap
ⓑ. Presence of d-orbitals for bonding expansion
ⓒ. Formation of highly stable ionic compounds only
ⓓ. Ability to expand octet beyond 8
Correct Answer: Formation of multiple bonds (C=C, N≡N, C=O) due to small size and strong p–p overlap
Explanation: Second-period elements (C, N, O) can form strong π-bonds due to effective overlap of small 2p orbitals. Heavier congeners (Si, P, S) have larger, diffuse orbitals, making π-bonding weaker.
This is the final section of Class 11 Chemistry MCQs – Chapter 3: Classification of Elements and Periodicity in Properties (Part 4). Based on the NCERT/CBSE syllabus, this chapter is extremely important for both board exams and competitive exams like JEE, NEET, and state-level tests. It covers advanced topics like periodic trends in properties, reasons for anomalous behavior of the first element, variable valency, different chemistry of heavier congeners, and the diagonal relationship. Across all 4 parts, you will find a total of 350 MCQs with step-by-step answers. In this last part, you will practice the final 50 questions with explanations, completing your preparation of this important chapter. These MCQs are perfect for final revision before exams and for sharpening your concepts for competitive tests.
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